1. ## Another Probability.

1) Four friends, 2 females and 2 males are playing contract bridge. Partners are randomly assigned. What is the probability that two females will be partners for the first game?

I did:
P(2 females) = P(no 2 females)
= P(6C2)
(8C2)
= 15
28

and i got it wrong....dont know what i did wrong.

2. Just list the simple events, there aren't that many of them.

3. Hello, questiongirl!

This is easier/harder than we'd expect . . .

Four friends, 2 females and 2 males are playing contract bridge.
Partners are randomly assigned.
What is the probability that two females will be partners for the first game?

Suppose the four friends are: . $\{F_1,\:F_2,\:M_1,\:M_2\}$

There are only three assignments for partners:

. . $\{F_1,F_2\:|\:M_1,M_2\}\quad \{F_1,M_1\:|\:F_2,\,M_2\} \quad \{F_1,\,M_2\:|\:F_2\,M_1\}$

Therefore: . $P(F_1,F_2) \;=\;\frac{1}{3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Choose a female.
. . Either one, it makes no difference.

Now what is the probability that her partner is the other female?

There are 3 choices: . $\text{other F},\;M_1,\;M_2$

Therefore: . $P(F_1F_2) \;=\;\frac{1}{3}$

4. Thanks Soroban...you see i have misinterpret the question. i thought it means that there are 4 friends, 2females, 2males so in total their are 8people.
okay i get it. thank you!